GMAT Prep Now OFFICIAL SOLUTION:

When x is divided by 11, the remainder is 5: So, the possible values of x are: 5, 16, 27, 38, etc.

When x is divided by 34, the remainder is 27: So, the possible values of x are: 27... STOP. Since both lists include 27, the smallest possible value of x is 27.

When y is divided by 17, the remainder is 11: So, the possible values of y are: 11, 28, 45, etc.

When y is divided by 3, the remainder is 2: So, the possible values of y are: 2, 5, 8, 11...STOP. Since both lists include 11, the smallest possible value of y is 11

Since the smallest possible values of x and y are 27 and 11 respectively, the smallest possible value of x + y is 38. So, C is the correct answer to the original question.

The Big Takeaway:

When solving remainder questions on the GMAT, you can sometimes save yourself a lot of work by listing possible values and, more importantly, by beginning with the smallest possible value.
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x/11= remainder is 5
x/34= remainder is 27
x=27

y/17= remainder is 11
y/3 = remainder is 2
y=11

x+y=38
Updated Answer to C , Thanks for correcting Engr2012
Cheers!!

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Your solution is not correct as shown below:

x= 11p+5 ---> x = 5 , 16, 27 ....
x= 34q+27 = 27, 61,...

Min common value of x =27

y=17r+11= 11,28,45...
y=3s+2=2,5,8,11...
Min y = 11

Thus least x+y value = 11+27=38

C is the correct answer.
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x=27 when divided by 11 leaves remainder 5, when divided by 34 leaves remainder 27.
y=11 when divided by 17leaves remainder 11, when divided by 3 leaves remainder 2

Hence x+y=38

Rule for finding the numbers are

Let x=34k+27 on division by 34 will leave remainder as 27

34k+27-5=34k+22 will be divisible by 11. For k=0.

Hence x=27.

Similarly Y can be found.

Answer : C
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For the least possible sum, lets assume that the quotient of all divisions are zero and that the given remainders are the actual numbers.
When x is divided by 11, the remainder is 5, and when x is divided by 34, the remainder is 27.
So the least value of X is 27.

When y is divided by 17, the remainder is 11, and when y is divided by 3, the remainder is 2.
So the least value of Y is 11.

Sum is 38

Hope my approach is correct

x= 11m+5 .... (1)
x=34n+27......(2)

From 1 and 2;
11m+5=34n+27
so, 11m-34n=22 ... (3)
for n=0, we get m=2, putting that in eq 1, we get x=22
for m=0, we get n=-22/7, we get -ve value of x, but we know x is a positive integer.

similarly, for y= 17p+11...(4)
and y= 3q+2 ....(5)
so, 17p-3q= -9 .....(6)
for, p=0, we get q=3 and y= 11 (from eq 5)
for q=0, we get p=-19/7 and a negative value of y.
but we know y is a positive integer.

so, x+y= 38
option c

this is my first ever post on gmat club Do give kudos if you feel i have helped in some way !
Cheers

Hi,

As it is mentioned when x is divided by 11, the remainder is 5, and when x is divided by 34, the remainder is 27. When y is divided by 17, the remainder is 11, and when y is divided by 3, the remainder is 2

Hence, I concluded when X is even (11+5) and Y is even (17+11); Even + Even = Even. Since there is just one option with an Even number. I picked C. Is this approach correct?

Solution to this problem is very easy. First, dont get scared by lengthy problem. Just look at Whats test maker is asking.
Test maker is asking to find out least possible value of x and y.
Always least value for multiples is 0.
Consider a multiple of 0 for values of x and y such that it satisfies given literature
For x, with zero multiple of largest number 34, remainder 27 is (34 x 0)+27 = 27
Similarly for y, with zero multiple of largest number 17, remainder 11 is (17 x 0) + 11 = 11
So x + y = 38.
Thats so simple

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